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Where should I put the variable scanner declaration? "int figureNumber = stdin.nextInt();"

I want to make it so that a user entering the wrong data type as figureNumber will see a message from me saying "Please enter an integer" instead of the normal error message, and will be given another chance to enter an integer. I started out trying to use try and catch, but I couldn't get it to work.
Sorry if this is a dumb question. It's my second week of an intro to java class.
import java. util.*;
public class Grades {
public static void main(String args []) {
Scanner stdin = new Scanner(System.in);
System.out.println();
System.out.print(" Please enter an integer: ");
int grade = stdin.nextInt();
method2 ();
if (grade % 2 == 0) {
grade -= 1;
}
for(int i = 1; i <=(grade/2); i++) {
method1 ();
method3 ();
}
}
}

public static void main(String args[]) {
Scanner stdin = new Scanner(System.in);
System.out.println();
System.out.print(" Welcome! Please enter the number of figures for your totem pole: ");
while (!stdin.hasNextInt()) {
System.out.print("That's not a number! Please enter a number: ");
stdin.next();
}
int figureNumber = stdin.nextInt();
eagle();
if (figureNumber % 2 == 0) { //determines if input number of figures is even
figureNumber -= 1;
}
for (int i = 1; i <= (figureNumber / 2); i++) {
whale();
human();
}
}
You need to check the input. The hasNextInt() method is true if the input is an integer. So this while loop asks the user to enter a number until the input is a number. Calling next() method is important because it will remove the previous wrong input from the Scanner.

Scanner stdin = new Scanner(System.in);
try {
int figureNumber = stdin.nextInt();
eagle();
if (figureNumber % 2 == 0) { //determines if input number of figures is even
figureNumber -= 1;
}
for(int i = 1; i <=(figureNumber/2); i++) {
whale();
human();
}
}
catch (InputMismatchException e) {
System.out.print("Input must be an integer");
}
You probably want to do something like this. Don't forget to add import java.util.*; at the beginning of .java file.

You want something in the form:
Ask for input
If input incorrect, say so and go to step 1.
A good choice is:
Integer num = null; // define scope outside the loop
System.out.println("Please enter a number:"); // opening output, done once
do {
String str = scanner.nextLine(); // read anything
if (str.matches("[0-9]+")) // if it's all digits
num = Integer.parseInt(str);
else
System.out.println("That is not a number. Please try again:");
} while (num == null);
// if you get to here, num is a number for sure
A do while is a good choice because you always at least one iteration.
It's important to read the whole line as a String. If you try to read an int and one isn't there the call will explode.

You can actually test the value before you assign it. You don't need to do any matching.
...
int figureNumber = -1;
while (figureNumber < 0) {
System.out.print(" Welcome! Please enter the number of figures for your totem pole: ");
if (stdin.hasNextInt()){
figureNumber = stdin.nextInt(); //will loop again if <0
} else {
std.next(); //discard the token
System.out.println("Hey! That wasn't an integer! Try again!");
}
}
...

How to check if an int contains a letter

I am trying to validate my code by error checking. I want to make sure the integer people enter does not contain a letter or more.
Here is my code. I am supposed to solve this problem using a one dimensional array. I got the code working but I am having problems with adding the error checking in.
Any help would be appreciated. Thanks
public void getNumbers() {
Scanner keyboard = new Scanner(System.in);
int array[] = new int[5];
int count = 0;
int entered = 0;
int k = -1;
while (entered < array.length) {
System.out.print("Enter a number ");
int number = keyboard.nextInt();
if (10 <= number && number <= 100) {
boolean containsNumber = false;
entered++;
for (int i = 0; i < count; i++) {
if (number == array[i]) // i Or j
{
containsNumber = true;
}
}
if (!containsNumber) {
array[count] = number;
count++;
} else {
System.out.println(number + " has already been entered");
}
} else {
System.out.println("number must be between 10 and 100");
}
//what does %d do?
for (int j = 0; j < count; j++) {
System.out.printf("%d ", array[j]);
}
System.out.println();
}
}
}

I'm assuming that you would want your program to ask the user to re-enter a number if they do not input a number the first time. In this scenario you might want to try something along the lines of this:
Scanner sc = new Scanner(System.in);
System.out.println("Enter a number: ");
while(!sc.hasNextInt()) {
//print some error statement
sc.nextLine();
}
int number = sc.nextInt();
System.out.println("Number is: " + number); // to show the value of number
// continue using number however you wish
Since hasNextInt() returns a boolean determining whether or not the input is an Integer, the program will never leave the while-loop until the program can confirm that the user has entered an integer.

keyboard.nextInt() will throw a InputMismatchException if you input a String.
If you want to check whether Scanner has an integer to read, you can use keyboard.hasNextInt().
Alternatively, you can read the input as
String s = keyboard.next() which will take the input as a String, and then use s.matches(".*\\d+.*") to detect whether or not it is an integer.
UPDATE: To answer questions -
keyboard.hasNextInt() will return a boolean. So for example, after System.out.print("Enter a number"), you could have an if statement checking to see if keyboard can receive numerical input, ie. if(keyboard.hasNextInt). If this is true, that means the user has entered numerical input, and you could continue with sayingint number = keyboard.nextInt(). If it is false, you would know that the user input is non-numerical.

Try/Catch with while loop can't work [closed]

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Problem
Method asks user for an integer value.
Pass the value into a loop to ensure it is a positive number.
Add a in.nextline to ensure the value don't overrun into next line of code. -
Return integer value.
private static int getIntFromUser(String aa) {
int aaa = 0;
while (true && aaa <= 0) {
try {
System.out.println(aa + ": ");
aaa = in.nextInt();
if (aaa <= 0) {
System.out.println("Please enter a positive number.");
}
} catch (Exception e) {
System.out.println("Please enter an integer: ");
in.next();
}
}
in.next();
return aaa;
}
Question
Try/Catch with while loop can't work ?

Tested it myself, it returns aaa properly (after removing in.next() at the bottom of the method and replacing it with in.nextLine()). When you do a return call, you need to send it somewhere. Such as System.out or to a variable like int x = getUserFromInput("test: ");
public class Tester {
private static Scanner in = new Scanner(System.in);
private static int getIntFromUser(String aa) {
int aaa = 0;
while (true && aaa <= 0) {
try {
System.out.println(aa + ": ");
aaa = in.nextInt();
if (aaa <= 0) {
System.out.println("Please enter a positive number.");
}
} catch (Exception e) {
System.out.println("Please enter an integer: ");
in.next();
}
}
in.nextLine();
return aaa;
}
public static void main(String[] a) {
System.out.println(getIntFromUser("test123"));
}
}
Output

Why doesn't the following program work? [closed]

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I can't figure out why the following program doesn't work. Please help me where did I make a mistake. Thank you.
import java.util.Scanner;
public class LargestNumber {
public static void main(String[] args) {
int[] numbers = new int[100];
int largestNumber = 0;
System.out.println("Enter numbers. When you want to finish, type 'finish'.");
Scanner sc = new Scanner(System.in);
do {
if (sc.hasNextInt()) {
for (int counter = 0; counter < 10; counter++)
numbers[counter] = sc.nextInt();
}
if (!sc.hasNextInt() && !sc.hasNext("finish")) {
System.out.println("It's neither number nor 'finish'.");
}
}
while (!sc.hasNext("finish"));
for (int x : numbers) {
if (x > largestNumber) {
largestNumber = x;
}
}
System.out.println("The largest number is: " + largestNumber);
}
}

This part:
do {
if (sc.hasNextInt()) {
for (int counter = 0; counter < 10; counter++)
numbers[counter] = sc.nextInt();
}
if (!sc.hasNextInt() && !sc.hasNext("finish")) {
System.out.println("It's neither number nor 'finish'.");
}
}
while (!sc.hasNext("finish"));
makes no sense. You:
test if the next thing in the input is an integer;
attempt to parse ten next tokens, assuming that they are all integers;
at the eleventh token you check whether it's another integer or "finish";
throw an exception if it's neither;
repeat everything if it's not "finish".
What you should actually do is something much, much simpler:
check next token:
if it's "finish", you're done;
if it's an integer, parse it;
otherwise throw error;
repeat this for up to 100 times;
you are done accepting input. Proceed to processing it.

I think that Scanner is unnessecarily complicated and doesn't work to much of the time. Here's how to do it the old fashioned way:
public class LargestNumber {
public static void main(String[] args) {
int largestNumber=0;
System.out.println("Enter numbers. When you want to finish, type 'finish'.");
BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
String line;
while (!(line = r.readLine()).equals("finish")) {
int val = Integer.parseInt(line);
if (val > largestNumber)
largestNumber = val;
}
System.out.println("The largest number is: " + largestNumber);
}
}
For this, enter each number on a new line. I used a shorter algorithm here, which is to read one number, and if it is bigger than the maximum so far, the new number is the maximum so far

According to the doc:
Throws:
InputMismatchException - if the next token does not match the Integer regular expression, or is out of range
So i thing you could do this instead:
int counter = 0;
while (counter < numbers.length) {
if (sc.hasNextInt()) {
numbers[counter++] = sc.nextInt();
} else {
if (sc.hasNext("finish")) {
sc.close();
break;
} else {
System.out.println("It's neither number nor 'finish'.");
sc.next();
}
}
}
for (int x : numbers) {
if (x > largestNumber) {
largestNumber = x;
}
}
System.out.println("The largest number is: " + largestNumber);
hope that helps

Handling Exceptions in Java - How to give user another chance?

Hey guys so i've been trying to answer this question for hours:
Write a program that asks the user to input a set of floating-point values. When the
user enters a value that is not a number, give the user a second chance to enter the
value. After two chances, quit reading input. Add all correctly specified values and
print the sum when the user is done entering data. Use exception handling to detect
improper inputs.
I've tried a few different things but i always have the same problem. Once something that isn't a number is given as input, the program outputs the message prompting for another input however the chance is not given, that is to say after 1 incorrect input it prints that message and jumps straight to printing the sum. The best i could do is below i'm just not sure how to approach this problem. Any help is greatly appreciated.
import java.util.Scanner;
import java.util.InputMismatchException;
public class q6{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
boolean firstChance = true;
boolean secondChance = true;
double sum = 0;
while (secondChance){
try{
while (firstChance){
try{
System.out.print("Please enter a number: ");
double input = in.nextDouble();
sum = sum + input;
}
catch (InputMismatchException ex){
firstChance = false;
}
System.out.print("Please enter a number to continue or something else to terminate: ");
double input = in.nextDouble();
sum = sum + input;
firstChance = true;
}
}
catch (InputMismatchException e){
secondChance = false;
}
}
System.out.print("The sum of the entered values is " + sum);
}
}

i'm just not sure how to approach this problem
Pseudocode could be as follows:
BEGIN
MAX_INPUT = 2;
i = 0;
WHILE i < MAX_INPUT
TRY
num = GET_NUM();
CATCH
continue;
FINALLY
i++
END WHILE
END

Since the parsing of input to double is not successful, the scanner does not go past the given input. From javadoc Scanner
Scans the next token of the input as a double. This method will throw
InputMismatchException if the next token cannot be translated into a
valid double value. If the translation is successful, the scanner
advances past the input that matched.
Since the translation is not successful, it does not advance. So, in the catch block you could call in.next() to skip the token.

You could use an integer counter instead of the boolean variables firstChance and secondChance and do something like:
int attempts = 0;
while(attempts < 2) {
try {
//Get user input - possible exception point
//Print sum
} catch(InputMismatchException e) {
attempts++;
//Continue?
}
}

import java.util.*;
public class q6 {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
double inputNumber, sum = 0.0;
int correctCount = 0, wrongCount = 0;
while(wrongCount <=1) {
System.out.println("Please enter a numeric floating value:");
try {
inputNumber = in.nextDouble();
correctCount++;
sum += inputNumber;
if(correctCount >= 2)
break;
} catch(InputMismatchException e) {
wrongCount++;
in = new Scanner(System.in);
continue;
}
}
System.out.println(sum);
}
}

You need to break your while loop when you encounter a "bad" input. Then, you'll need to set firstChance to true again, so you can access the second while, you also will need a counter that counts the number of the attempts (I named it chances):
int chances = 0;
while (secondChance){
firstChance = true;
try{
while (firstChance){
try{
System.out.print("Please enter a number: ");
double input = in.nextDouble();
sum = sum + input;
}
catch (InputMismatchException ex){
chances ++;
in = new Scanner(System.in);
firstChance = false;
}
if(!firstChance && chances < 2)
break;
if(chances >= 2) {
System.out.print("Please enter a number to continue or something else to terminate: ");
double input = in.nextDouble();
sum = sum + input;
firstChance = true;
}
}
}
catch (InputMismatchException e){
secondChance = false;
}
}

As stated in my comment, in order to not deal with Scanner wrong read, it would be better to use Scanner#nextLine() and read the data as String, then try to parse it as double using Double#parseDouble(String) since this method already throws NumberFormatException and you can handle this error. Also, it would be better that your code could handle more than 1 request (if your teacher or somebody else asks to do this):
final int REQUEST_TIMES = 2;
double sum = 0;
for(int i = 1; i <= REQUEST_TIMES; i++) {
while (true) {
try {
if (i < REQUEST_TIMES) {
System.out.print("Please enter a number: ");
} else {
System.out.print("Please enter a number to continue or something else to terminate: ");
}
String stringInput = in.nextLine();
double input = Double.parseDouble(stringInput);
sum += input;
} catch (NumberFormatException nfe) {
System.out.println("You haven't entered a valid number.");
break;
}
}
}

The logic is to read the input once, if it is correct then display the sum, else read the input again and display the sum if the input is correct.
public class q6 {
static int trial = 0;
public static void main(String[] args) {
double sum = 0;
while (trial <= 2) {
Scanner in = new Scanner(System.in);
try {
trial++;
System.out.print("Please enter a number: ");
double input = in.nextDouble();
sum = sum + input;
trial=3;
} catch (InputMismatchException ex) {
trial++;
if (trial == 4){
System.out.println("You have entered wrong values twice");
}
}
}
if (trial <= 3){
System.out.print("The sum of the entered values is " + sum);
}
}
The above program is just a rough idea, you can improve this to adapt to your need.